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erica [24]
3 years ago
7

Cal 1 Question (Optimization)

Mathematics
1 answer:
Sedaia [141]3 years ago
6 0
The base of the box is square, so if the sides of the base are x and the box has height y, then the volume of the box is

V=x^2y

We are maximizing this subject to the constraint of surface area,

72=\underbrace{5x^2}_{\text{glass top}}+\underbrace{4xy}_{\text{lateral faces}}+\underbrace{x^2}_{\text{base}}
72=6x^2+4xy

Solving for y gives

y=\dfrac{72-6x^2}{4x}

and substituting into the volume equation gives

V=x^2\dfrac{72-6x^2}{4x}
V=18x-\dfrac32x^3

Differentiating, we get

V'=18-\dfrac92x^2

which has roots at x=\pm2, but we omit the negative root. So x=2.

This means we must have

y=\dfrac{72-6(2)^2}{4(2)}=6
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